cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A219643 Least inverse of A219642; a(n) = minimal i such that A219642(i) = n.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 8, 10, 13, 15, 18, 21, 23, 26, 29, 32, 34, 36, 39, 42, 45, 47, 50, 54, 55, 57, 60, 63, 66, 68, 71, 75, 78, 81, 84, 89, 91, 94, 97, 100, 102, 105, 109, 112, 115, 118, 123, 126, 130, 134, 138, 143, 144, 146, 149, 152, 155, 157, 160, 164, 167
Offset: 0

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Author

Antti Karttunen, Nov 24 2012

Keywords

Links

Crossrefs

Cf. A219645 for the greatest inverse. A219644 gives the first differences.
This sequence is based on Fibonacci number system (Zeckendorf expansion): A014417. For analogous sequence in binary system, see: A213708, for factorial number system: A219653.

A219645 Greatest inverse of A219642; a(n) = maximal i such that A219642(i) = n.

Original entry on oeis.org

0, 1, 2, 4, 6, 7, 9, 12, 14, 17, 20, 22, 25, 28, 31, 33, 35, 38, 41, 44, 46, 49, 53, 54, 56, 59, 62, 65, 67, 70, 74, 77, 80, 83, 88, 90, 93, 96, 99, 101, 104, 108, 111, 114, 117, 122, 125, 129, 133, 137, 142, 143, 145, 148, 151, 154, 156, 159, 163, 166, 169
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Links

Crossrefs

Cf. A219643 for the least inverse. A219644 gives the first differences.
This sequence is based on Fibonacci number system (Zeckendorf expansion): A014417. Analogous sequence for binary system: A173601, for factorial number system: A219655.

Formula

a(n) = A219643(n)+A219644(n)-1.

A255054 Run lengths in A255072.

Original entry on oeis.org

1, 2, 3, 1, 4, 3, 1, 4, 4, 4, 3, 1, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 7, 6, 7, 8, 5, 6, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4
Offset: 0

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Comments

Number of integers k which require exactly n steps to reach 0, when starting from k and iterating the map: x -> x - (number of runs in binary representation of x).

Examples

			0 is the only number reached from 0 in zero steps, thus a(0) = 1.
Both 1 and 2, in binary '1' and '10', when the number of runs (A005811) is subtracted from them, result zero: 1-1 = 2-2 = 0, and these are only such numbers where the zero is reached with one step, thus a(1) = 2.
For 3, 4 and 5, in binary '11', '100' and '101', subtracting the number of runs results 2 in all cases, thus two steps are requires to reach zero, and as there are no other such cases, a(2) = 3.
For 6, in binary '110', subtracting A005811 repeatedly gives -> 6-2 = 4, 4-2 = 2, 2-2 = 0, three steps in total, and as 6 is the only such number requiring three steps, a(3) = 1.

Links

Crossrefs

Cf. A005811, A236840, A255053, A255055, A255072, A255123, A255124, A255056.
Cf. A255059 (positions of odd terms), A255060 (positions of even terms), A255061 (apart from its second term 1, gives positions of ones here).
Analogous sequences: A086876, A219644, A219654.

Formula

a(n) = A255053(n+1) - A255053(n).
a(n) = 1 + A255055(n) - A255053(n).
Other identities. For all n >= 0:
a(n) = 1 + A255123(n) + A255124(n).

A219654 Run lengths in A219652.

Original entry on oeis.org

1, 1, 2, 2, 2, 4, 4, 4, 4, 2, 4, 4, 4, 4, 6, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 6, 8, 4, 2, 4, 4, 4, 4, 6, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 6, 8, 6, 8, 8, 6, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 6, 8, 6, 8, 8, 6, 8, 10, 6, 6, 6, 6, 6, 6, 6, 8, 6, 8, 8, 6, 8, 10, 8, 10, 12, 6
Offset: 0

Views

Author

Antti Karttunen, Nov 25 2012

Keywords

Comments

a(n) tells from how many starting values one can end to 0 in n steps, with the iterative process described in A219652 (if going around in 0->0 loop is disallowed).

Links

Crossrefs

a(n) = 1+(A219655(n)-A219653(n)). This sequence is based on Factorial number system: A007623. Analogous sequence for binary system: A086876, for Zeckendorf expansion: A219644. Cf. A219652, A219659, A219666.

Formula

a(n) = A219653(n+1)-A219653(n). (The first differences of A219653).
Showing 1-4 of 4 results.

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